--- COMMENTARY: New: $\textbf{Exercise 19.}$ Suppose $W$ is finite-dimensional and $T \in \mathcal{L}(V, W)$. Prove that $T$ is injective if and only if there exists $S \in \mathcal{L}(W, V)$ such that $ST$ is the identity operator on $V$. $\textbf{Solution 19.}$ First suppose $T$ is injective. Define $S_0 : \text{range } T \to V$ by $S_0(Tv) = v$; because $T$ is injective, each element of range $T$ can be represented in the form $Tv$ in only one way, so $S_0$ is well defined.$\textbf{Solution 19 continued:}$ As can easily be checked, $S_0$ is a linear map on range $T$. By Exercise 13 in Section 3A, $S_0$ can be extended to a linear map $S \in \mathcal{L}(W, V)$. If $v \in V$, then $(ST)v = S(Tv) = S_0(Tv) = v$. Thus $ST$ is the identity operator on $V$, as desired. To prove the implication in the other direction, now suppose there exists $S \in \mathcal{L}(W, V)$ such that $ST$ is the identity operator on $V$. If $u, v \in V$ are such that $Tu = Tv$, then $u = (ST)(u) = S(Tu) = S(Tv) = (ST)v = v.$ Hence $u = v$. Thus $T$ is injective, as desired. --- $\textbf{Exercise 19 Commentary:}$ This shows that a linear map $T : V \to W$ is injective if and only if there exists a linear "inverse" map $S : W \to V$ such that $ST$ is the identity on $V$. This characterizes injectivity in terms of the existence of a right inverse. It is analogous to the fact that a number $a$ is invertible iff there exists $b$ such that $ab = 1$. These inverse maps play an important role in analyzing linear transformations and their ranges/nullspaces. $\textbf{Exercise 19 Examples:}$ 1. Let $T : \mathbb{R}^2 \to \mathbb{R}^3$ be the linear map given by $T(x,y) = (x, 2y, 3x-y)$. Then $T$ is injective, and a right inverse is given by $S(a,b,c) = (a, (c+b)/3)$. 2. Let $V = \mathcal{P}_2(\mathbb{R})$ and $W = \mathbb{R}^3$, and define $T : V \to W$ by $T(a + bx + cx^2) = (a, b, c)$. Then $T$ is injective, with right inverse $S(x,y,z) = x + yz + z^2$. 3. Let $V = \mathbb{C}^4$ and $W = \mathbb{C}^3$, and let $T : V \to W$ be the linear map with $4 \times 3$ matrix $\begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0 \end{pmatrix}$. Then $T$ is not injectiive, and there does not exist $S: W \to V$ with $ST = I_V$.